Derivation of Spring Deflection

Question

Derive an equation for the deflection of a spring in terms of its Young's modulus and dimensions.

Solution

The deflection of a spring can be derived using Hooke's Law and the properties of materials under stress and strain. The key assumptions are:

The spring is made from a wire with a circular cross-section.
The spring undergoes small deflections (linear elastic behavior).

Step 1: Hooke’s Law for Springs

Hooke's Law states that the force \( F \) applied to the spring is proportional to its deflection \( \delta \):

\( F = k \delta \)

Step 2: Relation Between Spring Constant and Material Properties

The stiffness \( k \) of a helical spring is given by:

\( k = \frac{G d^4}{8 D^3 n} \)

Step 3: Relating Shear Modulus \( G \) to Young’s Modulus \( E \)

The shear modulus \( G \) is related to Young’s modulus \( E \) as:

\( G = \frac{E}{2(1 + \nu)} \)

Step 4: Deflection \( \delta \) in Terms of Force and Material Properties

Substituting the expression for \( k \) into Hooke’s Law:

\( \delta = \frac{F}{k} = \frac{8 F D^3 n}{G d^4} \)

Step 5: Substitute \( G \) in Terms of \( E \)

Therefore, the deflection \( \delta \) is given by:

\( \delta = \frac{16 F D^3 n (1 + \nu)}{E d^4} \)

Final Equation for Deflection

Thus, the deflection \( \delta \) of the spring in terms of the applied force \( F \), the spring's dimensions, and the material's Young's modulus \( E \) is:

\( \delta = \frac{16 F D^3 n (1 + \nu)}{E d^4} \)